Numerical Examples.
553
EC
H. P. given out by generator = —7 =
H. P. Turbine must give to dynamo =
R of circuit, taking lead and return
(dia. of wire = "192, area = "03,
7=2640 ft.)
« -n i - -
H.P. lost in wire =
746
300 x 50
746
20
•"86
2640 x 8*4
•03 x 1,000,000
jjo_xjo x 74
746 ~
= 20
= 23-25
74 ohms
2-48
H. P. delivered to motor
is that generated less
= 20-2*48 = I7'52
that lost in wire.
H. P. available at shop shafting = 17*52 x '84
But H. P. given by turbine was 23*25
1471
„ ~ . H. P. taken out 14*71
Gross efficiency = —•--•--—- -. — = ——
--------------------- H.P.put in 23*25
= '6327 or 63! %
and efficiency of circuit only •
H.P. delivered to motor
H. P. generated
= ££2? = -876 oi*87*6°/0
there being 12*4% of the generated H. P. lost in the wire.
Two actual cases may be quoted (i) 4^ H. P. was trans-
mitted 8 miles through •£%" telegraph wire, with a total efficiency
of 30 %. (2) The dynamos having a resistance of 470 ohms, and
the circuit 950 ohms, the line being 34 miles long, a total effi-
ciency of 32 % was obtained by decreasing revolutions from 2100
at generator to 1400 at motor, the potentials dropping 2400 to
1600 volts, a method of working first advised by Siemens.
(Seep. 929.)
Storage cells are objectionable for tramcar and locomotive
driving on account of their great weight, 2 tons of cells being
about the weight required for a i-ton car. The following results
are from an actual experiment with Faure accumulators :
35 cells of 95 Ibs. each
H. P. absorbed in charging
Time of charging
Lost work in charging
Chemically stored energy
Recovered electric energy
tons.
2 2 hrs. 45 mins.
34%
66%
60 % of 66 % = ,